Circles

Introduction

Chapter 10: Circles

Page 1 of 5: Introduction

Welcome to the fascinating world of circles! You've known them since you were little, but there's so much more to discover. A circle is simply a collection of all points in a plane that are at a fixed distance (the radius) from a fixed point (the centre).

In this lesson, we explore the relationship between a circle and a straight line. Imagine a train track (a straight line) running past a circular garden. The track might be far away, cut through a part of the garden, or just graze its edge. These three scenarios form the basis of our study. We will introduce three key terms: non-intersecting line, secant, and the star of this chapter, the tangent. Understanding these will unlock powerful properties and theorems that have applications in everything from engineering and physics to art and design.

{{VISUAL: diagram: A single circle with center O. Three separate lines are shown in the same plane. Line 1 (PQ) does not touch the circle. Line 2 (AB) cuts through the circle at two points, X and Y. Line 3 (MN) touches the circle at exactly one point, T.}}

Definitions & Key Terms

Before we dive deep, let's get our vocabulary straight. These terms are the building blocks for the entire chapter.

Derivation: Why is a Tangent Perpendicular to the Radius?

This is the first and most fundamental theorem of this chapter. It's a simple yet powerful idea that forms the basis for many problems.

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Let's break down the logic step-by-step.

{{VISUAL: diagram: A circle with center O. A tangent line XY touches the circle at point P. The radius OP is drawn. Another point Q is marked on the line XY (not at P) and a line segment OQ is drawn connecting the center to this point. OQ is clearly longer than OP.}}

Proof of Theorem 10.1

1. Setup: We start with a circle with centre O and a tangent line XY that touches the circle at the point of contact, P. The line segment OP is the radius through the point of contact. Our goal is to prove that OP is perpendicular to XY (i.e., ∠OPX = 90°).

2. Choose a Point: Let's take any point Q on the tangent line XY, other than P. Join O and Q with a line segment.

3. Analyze the Position of Q: Since P is the only point the tangent has in common with the circle, the point Q must lie outside the circle. If Q were inside, the line XY would have to cross the circle again to get out, making it a secant, not a tangent.

4. Compare Distances: Because Q is outside the circle, the distance from the centre to Q (OQ) must be greater than the distance from the centre to the edge of the circle (the radius, OP).

   OQ > OP
   

5. Generalize the Finding: This logic holds true for every single point on the line XY except for P. No matter where you place Q on the tangent line, the distance OQ will always be longer than OP.

6. The Shortest Distance: This means that OP is the shortest distance from the centre O to any point on the line XY. We know from geometry that the shortest distance from a point to a line is the perpendicular distance. Therefore, OP must be perpendicular to XY.

   OP ⊥ XY
   

{{KEY: type=theorem | title=Theorem 10.1: Radius-Tangent Perpendicularity | text=The radius drawn to the point of contact of a tangent is always perpendicular (at a 90° angle) to the tangent line.}}

Solved Examples

Let's apply these concepts to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Identifying Lines (Easy)

Given: A circle with centre O and several lines: AB, CD, PQ, and RS as shown in the figure.

To Find: Identify the tangent, secant, and non-intersecting line from the figure.

Solution:

1. Examine line AB: The line AB passes through the circle, intersecting it at two distinct points. Therefore, it is a secant.

2. Examine line CD: The line CD touches the circle at exactly one point, P. Therefore, it is a tangent. The point P is the point of contact.

3. Examine line RS: The line RS does not touch or intersect the circle at any point. Therefore, it is a non-intersecting line.

Final Answer: Tangent: CD, Secant: AB, Non-intersecting line: RS

Example 2: Finding Tangent Length (Medium)

Given: A point P is 13 cm from the centre O of a circle. The length of the tangent drawn from P to the circle is 12 cm.

To Find: The radius of the circle.

Solution:

1. Let the point of contact on the circle be T. We are given OP = 13 cm and the tangent length PT = 12 cm. We need to find the radius, OT.

2. According to Theorem 10.1, the radius through the point of contact is perpendicular to the tangent. So, OT ⊥ PT.

3. This means that ΔOTP is a right-angled triangle with the right angle at T. The hypotenuse is the side opposite the right angle, which is OP.

4. Using the Pythagorean theorem (a² + b² = c²):

   OT² + PT² = OP²
   

5. Substitute the given values:

   OT² + 12² = 13²
   

   OT² + 144 = 169
   

6. Solve for OT:

   OT² = 169 - 144
   

   OT² = 25
   

   OT = √25 = 5 cm
   

Final Answer: The radius of the circle is 5 cm.

Example 3: Concentric Circles (Hard)

Given: Two concentric circles (circles with the same center) have radii 5 cm and 3 cm. A chord of the larger circle touches the smaller circle.

To Find: The length of this chord.

{{VISUAL: diagram: Two concentric circles with common center O. The smaller circle has radius 3cm (OP) and the larger has radius 5cm (OA). A horizontal chord AB of the larger circle is drawn such that it is tangent to the smaller circle at point P. Triangle OPA is a right-angled triangle.}}

Solution:

1. Let the common center be O. Let the chord of the larger circle be AB, which touches the smaller circle at point P.

2. OA is the radius of the larger circle, and OP is the radius of the smaller circle. Thus, OA = 5 cm and OP = 3 cm.

3. Since AB is a tangent to the smaller circle at point P, by Theorem 10.1, the radius OP must be perpendicular to the tangent AB.

   OP ⊥ AB
   

4. This means ΔOPA is a right-angled triangle with the right angle at P. OA is the hypotenuse.

5. Using the Pythagorean theorem in ΔOPA:

   OP² + AP² = OA²
   

   3² + AP² = 5²
   

   9 + AP² = 25
   

   AP² = 16
   

   AP = 4 cm
   

6. In the larger circle, OP is a perpendicular from the center to the chord AB. A key property of circles (from Class IX) is that a perpendicular from the center to a chord bisects the chord. Therefore, P is the midpoint of AB.

   AB = 2 × AP
   

   AB = 2 × 4 = 8 cm
   

Final Answer: The length of the chord is 8 cm.

Example 4: Number of Parallel Tangents (Tricky)

Given: A circle and a secant PQ.

To Find: What is the maximum number of tangents that can be drawn to the circle parallel to the secant PQ?

Solution:

1. Visualize a secant PQ cutting through a circle. Now, imagine drawing lines parallel to PQ.

2. If we draw parallel lines on one side of PQ, they will also be secants, but the chords they form will get progressively shorter.

3. Eventually, there will be one specific parallel line that just touches the circle at a single point. This is a tangent. If we move any further, the lines will become non-intersecting.

4. Similarly, if we draw parallel lines on the other side of the secant PQ, the same process will occur. The chords will shorten until one line just touches the circle, forming another tangent.

5. These two tangents, one on each side of the secant, are at the "extreme" positions. Any other line parallel to PQ will either be a secant (if it's between the two tangents) or a non-intersecting line (if it's outside them).

6. Therefore, for any given secant, there are exactly two tangents that can be drawn parallel to it.

Final Answer: A maximum of two tangents can be drawn parallel to a given secant.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Many students make these simple errors. Be careful to avoid them!

Brain-Teaser Questions

Ready for a challenge? Test your understanding with these higher-order thinking questions.

1. A line is drawn at a distance of 8 cm from the center of a circle with a radius of 6 cm. How many points does this line have in common with the circle?

   💡 Answer: Zero. The distance of the line from the center (8 cm) is greater than the radius (6 cm). This means the line is entirely outside the circle and is a non-intersecting line.

2. If two tangents are drawn to a circle from an external point, are they always equal in length? (Think visually, you don't need a formal proof yet).

   💡 Answer: Yes. If you draw a diagram, you can see two right-angled triangles formed by the tangents, the radii to the points of contact, and the line joining the center to the external point. These triangles look congruent, which would imply the tangent lengths are equal. We will prove this later!

3. What is the maximum number of common tangents that can be drawn to two circles that do not touch or intersect each other?

   💡 Answer: Four. You can draw two "direct" common tangents (like the belts on two pulleys) and two "transverse" common tangents that cross each other in the space between the circles.

Mini Cheatsheet

Screenshot this table for your last-minute revision!

Tangent to a Circle — Definition and Existence

Page 2: Tangent to a Circle — Definition and Existence

Concept Introduction

Imagine a train moving on a straight track, and next to it, a giant Ferris wheel is spinning. From the train's perspective, for a fleeting moment, one of the cabins on the wheel seems to be moving perfectly parallel to the train's path, just "touching" its line of sight. This line of sight represents a tangent.

In geometry, a tangent is a straight line that "just touches" a curve at a single point without crossing it. The word itself comes from the Latin tangere, meaning "to touch." For circles, this concept is fundamental. The ground beneath a rolling bicycle wheel acts as a tangent to the wheel at the point of contact. This single point of contact has a very special relationship with the center of the circle, which forms the basis of many geometric properties and real-world applications, from gear systems to satellite orbits.

{{FORMULA: expr=OP ⊥ XY | symbols=O:Center of the circle, P:Point of contact, XY:Tangent line}}

Definitions & Key Terms

Here are the essential terms you'll need to understand for this topic.

The Relationship Between a Tangent and a Radius

One of the most powerful properties in the study of circles is the relationship between a tangent and the radius at the point of contact. Let's explore the logic behind this core theorem.

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

This theorem states that if you draw a radius to the point where a tangent touches the circle, the angle between the radius and the tangent will always be exactly 90°.

Proof of Theorem 10.1

Let's break down the proof logically, step-by-step.

Given: A circle with center O and a tangent line XY that touches the circle at point P.

To Prove: The radius OP is perpendicular to the tangent XY (i.e., OP ⊥ XY).

{{VISUAL: diagram: A circle with center O. A tangent line XY touches the circle at point P. The radius OP is shown with a right-angle symbol at P. Another point Q is marked on the line XY, and a dotted line segment OQ is drawn connecting the center to Q.}}

Steps:

1. Consider a point Q on the tangent line. Let's take any point Q on the line XY, other than the point of contact P. Join the center O to this point Q.

2. Determine the location of point Q. Since P is the only point on the line XY that is also on the circle (by the definition of a tangent), the point Q must lie outside the circle. If Q were inside, the line XY would intersect the circle at two points, making it a secant, not a tangent.

3. Compare the lengths of OQ and OP. Because Q is outside the circle, the distance from the center O to Q (the length of OQ) must be greater than the distance from the center O to any point on the circle (the radius). The segment OP is a radius. Therefore:

   OQ > OP
   

4. Generalize for all points on the tangent. This condition (OQ > OP) is true for every single point on the line XY except for the point P itself. This means that OP is the shortest of all possible distances from the point O to the line XY.

5. Conclusion. A fundamental property of geometry (Theorem A1.7 in your textbook appendix) states that the shortest distance from a point to a line is the perpendicular distance. Since OP is the shortest distance from O to XY, it must be perpendicular to XY.

   OP ⊥ XY
   

   Hence, the theorem is proved.

{{KEY: type=concept | title=The Shortest Distance is Always Perpendicular | text=The entire proof of Theorem 10.1 hinges on a simple, intuitive idea: the shortest path from a point to a straight line is the one that meets the line at a right angle. Since the radius to the point of contact is the shortest possible line from the center to the tangent, it must be perpendicular.}}

Solved Examples

Let's apply these concepts to solve some problems, ranging from easy to tricky.

Example 1: Finding an Angle

Given: A circle with center O. PQ is a tangent from point P. If ∠OQP = 30° and the radius is 7 cm.

To Find: The measure of ∠QOP.

Solution:

1. Identify the point of contact and the radius. The line PQ is tangent at point P. Therefore, OP is the radius to the point of contact.

2. Apply Theorem 10.1. The radius at the point of contact is perpendicular to the tangent. This means ΔOPQ is a right-angled triangle.

   ∠OPQ = 90°
   

3. Use the angle sum property of a triangle. The sum of angles in ΔOPQ is 180°.

   ∠OPQ + ∠OQP + ∠QOP = 180°
   

4. Substitute the known values and solve for ∠QOP.

   90° + 30° + ∠QOP = 180°
   

   120° + ∠QOP = 180°
   

   ∠QOP = 180° - 120° = 60°
   

Final Answer: The measure of ∠QOP is 60°.

Example 2: Finding the Length of a Tangent (Medium)

Given: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm.

To Find: The length of the tangent PQ.

{{VISUAL: diagram: A right-angled triangle OPQ, with the right angle at P. O is the center of a circle with radius OP = 5 cm. P is the point of contact on the circle. Q is an external point on the tangent line. The hypotenuse OQ is labeled 12 cm. The length of the tangent, PQ, is marked with a question mark.}}

Solution:

1. Draw a diagram representing the information. The line PQ is tangent at P, and OP is the radius.

2. According to Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. Therefore, OP ⊥ PQ. This makes ΔOPQ a right-angled triangle with the right angle at P.

3. In the right-angled ΔOPQ, the side opposite the right angle, OQ, is the hypotenuse. We can apply the Pythagorean theorem: (Perpendicular)² + (Base)² = (Hypotenuse)².

   OP² + PQ² = OQ²
   

4. Substitute the given values: OP = 5 cm and OQ = 12 cm.

   5² + PQ² = 12²
   

   25 + PQ² = 144
   

5. Solve for PQ².

   PQ² = 144 - 25
   

   PQ² = 119
   

6. Find the length of PQ by taking the square root.

   PQ = √119 cm
   

Final Answer: The length of PQ is √119 cm.

Example 3: Chord of Concentric Circles (Hard)

Given: Two concentric circles (circles with the same center O) have radii of 5 cm and 3 cm. A chord of the larger circle touches the smaller circle.

To Find: The length of this chord.

{{VISUAL: diagram: Two concentric circles with center O. The larger circle has a chord AB. This chord AB is tangent to the smaller circle at point P. The radius of the smaller circle, OP, is shown and labeled 3 cm. The radius of the larger circle, OA, is also drawn and labeled 5 cm, forming a right-angled triangle OPA.}}

Solution:

1. Let the larger circle be C₁ and the smaller circle be C₂. Let AB be the chord of C₁ that is tangent to C₂ at point P. O is the common center.

2. Since AB is tangent to the smaller circle C₂ at point P, the radius OP of C₂ is perpendicular to the chord AB.

   OP ⊥ AB
   

3. Now, consider the chord AB of the larger circle C₁. We know that a perpendicular drawn from the center of a circle to a chord bisects the chord. Since OP ⊥ AB, P must be the midpoint of AB.

   AP = PB
   

4. Join O to A. Now consider the triangle ΔOPA. It is a right-angled triangle with the right angle at P.

   • OA is the radius of the larger circle = 5 cm (Hypotenuse).
   • OP is the radius of the smaller circle = 3 cm (Perpendicular).
   • AP is half the length of the chord (Base).

5. Apply the Pythagorean theorem to ΔOPA.

   OP² + AP² = OA²
   

   3² + AP² = 5²
   

   9 + AP² = 25
   

   AP² = 25 - 9 = 16
   

   AP = √16 = 4 cm
   

6. Since P is the midpoint of AB, the full length of the chord is twice the length of AP.

   AB = 2 × AP = 2 × 4 = 8 cm
   

Final Answer: The length of the chord is 8 cm.

Example 4: Number of Tangents (Tricky Application)

Given: A circle and a line L on a plane.

To Find: The maximum number of tangents that can be drawn to the circle which are parallel to the line L.

Solution:

1. This is a conceptual problem based on Activity 2 in your textbook. Imagine a secant PQ passing through the circle, parallel to the given line L.

2. If we draw other lines parallel to PQ (and thus parallel to L), the length of the chord formed by the intersection of the line and the circle changes.

3. As we move the parallel line away from the center, the chord length decreases. Eventually, the two points of intersection merge into a single point. At this stage, the secant becomes a tangent.

4. This can happen on both sides of the center. There will be one position on one side of the center where the line becomes a tangent, and one position on the other side.

5. These two tangent positions represent the "extreme" parallel lines that still touch the circle. Any line further away will not intersect the circle at all.

6. Therefore, there are exactly two tangents to the circle that can be parallel to a given line. These two tangents touch the circle at opposite ends of a diameter that is perpendicular to the given line L.

Final Answer: A circle can have 2 parallel tangents at the most.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Prove that the perpendicular drawn at the point of contact to the tangent of a circle must pass through the center of the circle. (This is the converse of Theorem 10.1).

   💡 Answer: Let XY be the tangent at point P to a circle with center O. Let's assume the perpendicular to XY at P does not pass through O. Let it pass through another point, O'. So, O'P ⊥ XY. But we know from Theorem 10.1 that the radius OP is also perpendicular to XY (OP ⊥ XY). This means both OP and O'P are perpendicular to the same line XY at the same point P. This is only possible if O'P and OP are the same line, which implies that O' must lie on the line OP. This contradicts our assumption unless O' is the same as O. Therefore, the perpendicular at the point of contact must pass through the center.

2. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. If OP is the radius and OT intersects the circle at point R, and OR = RT, what is the measure of ∠OTP?

   💡 Answer: In right-angled ΔOPT (since OP ⊥ PT), we have OT = OR + RT. Given OR = RT and OR is the radius (like OP), we have OT = OP + OP = 2 × OP. Now, in ΔOPT, we can use trigonometry. sin(∠OTP) = Opposite/Hypotenuse = OP/OT = OP / (2 × OP) = 1/2. The angle whose sine is 1/2 is 30°. So, ∠OTP = 30°.

3. From a point P, which is at a distance of 13 cm from the center O of a circle of radius 5 cm, a pair of tangents PQ and PR are drawn to the circle. What is the area of the quadrilateral PQOR?

   💡 Answer: The quadrilateral PQOR is formed by two right-angled triangles, ΔPQO and ΔPRO (right-angled at Q and R respectively). In ΔPQO, by Pythagoras theorem: OQ² + PQ² = OP² → 5² + PQ² = 13² → 25 + PQ² = 169 → PQ² = 144 → PQ = 12 cm. The area of ΔPQO = ½ × base × height = ½ × PQ × OQ = ½ × 12 × 5 = 30 cm². Since ΔPQO and ΔPRO are congruent (by RHS congruence), the area of quadrilateral PQOR is 2 × Area(ΔPQO). Area(PQOR) = 2 × 30 = 60 cm².

Mini Cheatsheet

Tangent to a Circle — Properties (Theorem 10.1)

Tangent to a Circle — Properties (Theorem 10.1)

Welcome to the next step in our journey through circles! We've seen what a tangent is — a line that just kisses the circle at a single point. But what happens at that exact point of contact? Is there a special relationship between the tangent and the circle's radius? This page uncovers a fundamental property that forms the basis for almost every problem involving tangents.

Imagine a bicycle wheel rolling perfectly straight on the ground. The ground acts as a tangent to the circular wheel. At any moment, only one point on the wheel touches the ground. The spoke connecting the wheel's center to this point of contact seems to stand perfectly upright, forming a 90° angle with the ground. This real-world observation is a beautiful demonstration of a core mathematical theorem we are about to prove and master.

{{FORMULA: expr=OP² + PQ² = OQ² | symbols=OP:Radius, PQ:Tangent Length, OQ:Distance from center to external point}}

Definitions & Key Terms

Before we dive into the proof, let's solidify our understanding of the key terms involved. These concepts are the building blocks for everything that follows.

The Core Property: Radius is Perpendicular to the Tangent

This is the most important theorem for this section. It establishes a fixed, predictable relationship between a circle's radius and its tangent.

{{KEY: type=theorem | title=Theorem 10.1 | text=The tangent at any point of a circle is perpendicular to the radius through the point of contact.}}

This means that wherever you draw a tangent, if you then draw a radius to that exact point of contact, the angle between them will always be 90°. This allows us to use properties of right-angled triangles, especially the Pythagorean theorem, to solve problems.

Proof of Theorem 10.1

Let's walk through the logical proof, just as a mathematician would. The strategy here is to show that the radius is the shortest possible line from the center to the tangent, and the shortest distance from a point to a line is always the perpendicular distance.

{{VISUAL: diagram: A circle with center O. A tangent line XY touches the circle at point P. A radius OP is drawn. Another point Q is marked on the line XY, and a line segment OQ is drawn. The diagram shows OQ > OP.}}

Given: A circle with center O and a tangent XY which touches the circle at point P.

To Prove: The radius OP is perpendicular to the tangent XY (i.e., OP ⊥ XY).

Proof Steps:

1. Let's take any point Q on the tangent line XY, other than the point of contact P. Join the center O to this point Q.

2. Now, consider the location of point Q. Since P is the only point where the line XY touches the circle, the point Q must lie outside the circle. If it were inside, the line XY would become a secant, intersecting the circle at two points, which contradicts our definition of a tangent.

3. Because Q is outside the circle, the length of the line segment OQ must be greater than the length of the radius OP.

   OQ > OP
   

4. This condition is true for every single point on the line XY except for the point P itself. You can pick any other point on the tangent, and the line connecting it to the center will always be longer than the radius.

5. This means that OP is the shortest of all possible distances from the center O to any point on the line XY.

6. A fundamental property in geometry states that the shortest distance from a point to a line is the perpendicular line segment. Therefore, OP must be perpendicular to XY.

   OP ⊥ XY
   

This elegant proof confirms our bicycle wheel observation and gives us a powerful tool for problem-solving.

Solved Examples

Let's apply Theorem 10.1 to solve some typical problems. Notice how each solution hinges on finding a right-angled triangle.

Example 1: Basic Angle Property

Given: A circle with center C. A line L is tangent to the circle at point T. A radius CT is drawn.

To Find: The measure of the angle ∠CTL.

Solution:

1. According to Theorem 10.1, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. Here, L is the tangent, T is the point of contact, and CT is the radius through the point of contact.

3. Therefore, the radius CT is perpendicular to the tangent L.

   ∠CTL = 90°
   

Final Answer:

The measure of the angle ∠CTL is 90°.

Example 2: Finding Tangent Length (NCERT Problem)

Given: A tangent PQ at a point P of a circle of radius 5 cm. This tangent meets a line through the centre O at a point Q such that OQ = 12 cm.

To Find: The length of the tangent PQ.

{{VISUAL: diagram: A right-angled triangle OPQ. O is the center of a circle. P is on the circle (point of contact). Q is a point outside the circle. OP is the radius = 5 cm. OQ is the hypotenuse = 12 cm. PQ is the tangent. The angle at P is marked as 90°.}}

Solution:

1. We know that the radius OP is perpendicular to the tangent PQ at the point of contact P. Therefore, ΔOPQ is a right-angled triangle with the right angle at P.

2. In a right-angled triangle, the side opposite the right angle is the hypotenuse. Here, OQ is the hypotenuse.

3. We can apply the Pythagorean theorem: (Perpendicular)² + (Base)² = (Hypotenuse)². In ΔOPQ, this translates to OP² + PQ² = OQ².

4. Substitute the given values: OP = 5 cm and OQ = 12 cm.

   5² + PQ² = 12²
   

5. Calculate the squares.

   25 + PQ² = 144
   

6. Isolate PQ² by subtracting 25 from both sides.

   PQ² = 144 - 25
   

   PQ² = 119
   

7. Take the square root to find the length of PQ.

   PQ = √119 cm
   

Final Answer:

The length of PQ is √119 cm.

Example 3: Finding the Radius

Given: A point A is 13 cm away from the center O of a circle. The length of the tangent drawn from A to the circle at point P is 12 cm.

To Find: The radius of the circle.

Solution:

1. The radius OP is perpendicular to the tangent AP at the point of contact P. This means ΔOPA is a right-angled triangle, with the right angle at P.

2. The hypotenuse is the side opposite the right angle, which is OA.

3. Apply the Pythagorean theorem: OP² + AP² = OA².

4. Substitute the given values: AP = 12 cm and OA = 13 cm.

   OP² + 12² = 13²
   

5. Calculate the squares.

   OP² + 144 = 169
   

6. Isolate OP² by subtracting 144 from both sides.

   OP² = 169 - 144
   

   OP² = 25
   

7. Take the square root to find the length of the radius OP.

   OP = √25 = 5 cm
   

Final Answer:

The radius of the circle is 5 cm.

Example 4: Parallel Tangents and Diameter

Given: A circle has two parallel tangents, XY and X'Y'. The distance between these two tangents is 16 cm.

To Find: The radius of the circle.

{{VISUAL: diagram: A circle with center O. A horizontal tangent XY is at the top, touching at point P. A parallel horizontal tangent X'Y' is at the bottom, touching at point Q. A vertical line segment PQ passes through the center O, representing the diameter.}}

Solution:

1. Let the points of contact for the parallel tangents XY and X'Y' be P and Q respectively. The distance between the parallel tangents is the length of the perpendicular line segment connecting them.

2. We know that the radius OP is perpendicular to tangent XY (OP ⊥ XY), and radius OQ is perpendicular to tangent X'Y' (OQ ⊥ X'Y').

3. Since XY is parallel to X'Y', the perpendiculars to these lines (OP and OQ) must lie on the same straight line. This means that POQ is a straight line segment.

4. The line segment POQ passes through the center O and connects two points on the circle, making it the diameter of the circle.

5. The distance between the parallel tangents is therefore equal to the length of the diameter.

   Diameter = PQ = 16 cm
   

6. The radius is half of the diameter.

   Radius = Diameter / 2 = 16 / 2
   

   Radius = 8 cm
   

Final Answer:

The radius of the circle is 8 cm.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Here are some common pitfalls to avoid. Seeing the wrong way helps you remember the right way!

Brain-Teaser Questions

Ready to test your understanding at a deeper level?

1. A line segment AB is tangent to a circle with center O at point P. If we say the "normal" to the circle at P passes through a point Q, and Q is 20 cm away from P, what can you say about the location of the center O?

   💡 Answer: The 'normal' is the line containing the radius at the point of contact. Since the radius always passes through the center O, the normal must also pass through O. Therefore, the center O must lie on the line segment PQ.

2. A point P is on a circle with center O and radius r. A tangent is drawn at P. Point Q lies on this tangent such that the area of ΔOPQ is 24 cm² and the length of the tangent PQ is 8 cm. What is the radius of the circle?

   💡 Answer: We know OP ⊥ PQ, so ΔOPQ is a right-angled triangle. The area of a triangle is ½ × base × height. We can take PQ as the base and OP (the radius) as the height. Area = ½ × PQ × OP 24 = ½ × 8 × r 24 = 4r r = 24 / 4 = 6 cm. The radius is 6 cm.

3. Consider a circle. Can you draw two tangents that are perpendicular to each other and also parallel to each other? Explain your reasoning.

   💡 Answer: You cannot. Two lines can be either parallel or perpendicular (or neither), but not both simultaneously. Parallel lines never meet, while perpendicular lines intersect at a 90° angle. Therefore, it's impossible for two tangents to be both parallel and perpendicular to each other.

Mini Cheatsheet

Screenshot this table for a quick revision of all the key concepts from this page!

Number of Tangents from a Point on a Circle

Number of Tangents from a Point to a Circle

Welcome back! In the previous section, we established the fundamental relationship between a tangent and the radius at the point of contact. Now, let's explore a fascinating question: How many tangents can you draw to a circle from a single point? Does it matter where the point is located?

Imagine you are standing in a large circular field. If you are right in the center, you can look out in any direction and your line of sight will always stay within the field. If you walk to the very edge of the field, there's only one perfect line of sight that just skims the boundary. But what if you step outside the field? Suddenly, you can see two distinct "edges" of the circle. Your lines of sight to these edges are like tangents. This simple analogy captures the core idea we'll investigate mathematically: the position of the point determines the number of possible tangents.

{{FORMULA: expr=PQ = PR | symbols=P:an external point, Q:point of contact 1, R:point of contact 2}}

Key Concepts and Cases

Let's formalize our observations from the activity described in the textbook. The number of tangents you can draw from a point P to a circle depends entirely on whether P is inside, on, or outside the circle.

A crucial term we introduce here is the length of the tangent. This is defined as the length of the line segment from the external point to the point of contact on the circle. For instance, if P is outside and T is the point of contact, the length of the tangent is PT.

Theorem 10.2: The External Point Tangent Theorem

This is one of the most important theorems in this chapter. It states a beautiful and symmetric property of tangents drawn from an external point.

{{KEY: type=theorem | title=Theorem 10.2 | text=The lengths of tangents drawn from an external point to a circle are equal.}}

Let's walk through the proof step-by-step. It's an elegant application of triangle congruence.

Proof of Theorem 10.2

Given: A circle with center O. An external point P from which two tangents, PQ and PR, are drawn to the circle, touching it at points Q and R respectively.

To Prove: The lengths of the tangents are equal, i.e., PQ = PR.

{{VISUAL: diagram: A circle with center O. An external point P. Two tangents PQ and PR are drawn from P to the circle, touching at Q and R. Lines OQ, OR, and OP are drawn as dotted lines. Right angle symbols are marked at ∠OQP and ∠ORP.}}

Proof Steps:

1. Construction: To begin the proof, we connect the center O to the external point P, and also to the points of contact Q and R. This gives us two triangles: ΔOQP and ΔORP.

2. Identify Right Angles: We know from Theorem 10.1 that a tangent is perpendicular to the radius at the point of contact. Therefore, OQ ⊥ PQ and OR ⊥ PR. This means that ∠OQP and ∠ORP are both right angles (90°).

3. Analyze the Triangles: Now we have two right-angled triangles, ΔOQP and ΔORP. Let's compare their sides.

   • OQ = OR (Both are radii of the same circle).
   • OP = OP (This is the common side to both triangles).
   • ∠OQP = ∠ORP = 90° (As established in Step 2).

4. Prove Congruence: Based on the properties above, we can see that the two triangles are congruent by the RHS (Right angle - Hypotenuse - Side) congruence rule.

   ΔOQP ≅ ΔORP
   

5. Conclusion: Since the triangles are congruent, their corresponding parts must be equal. This is often abbreviated as CPCT (Corresponding Parts of Congruent Triangles). Therefore, the side PQ must be equal to the side PR.

   PQ = PR
   

This completes the proof. It's a simple yet powerful result that we will use extensively.

Remark: As noted in the textbook, this can also be proved using the Pythagoras Theorem. In ΔOQP, PQ² = OP² – OQ². In ΔORP, PR² = OP² – OR². Since OQ = OR (radii), it follows that PQ² = PR², which means PQ = PR.

Solved Examples

Let's apply these concepts to solve some problems, ranging from straightforward to more complex.

Example 1: Finding the Radius (Easy)

Given: The length of a tangent from a point Q to a circle is 24 cm. The distance of Q from the center of the circle is 25 cm.

To Find: The radius of the circle.

Solution:

1. Let the center of the circle be O and the point of contact be T. We are given QT = 24 cm and OQ = 25 cm.

2. According to Theorem 10.1, the radius OT is perpendicular to the tangent QT at the point of contact T. Therefore, ΔOTQ is a right-angled triangle with the right angle at T.

3. The side OQ is opposite the right angle, making it the hypotenuse. We can apply the Pythagoras theorem: OT² + QT² = OQ².

4. Substitute the given values into the equation. Let r be the radius (OT).

   r² + 24² = 25²
   

5. Solve for r².

   r² + 576 = 625
   

   r² = 625 - 576
   

   r² = 49
   

6. Take the square root to find the radius. Length cannot be negative.

   r = √49 = 7
   

Final Answer: The radius of the circle is 7 cm.

Example 2: Chord of a Larger Circle (Medium)

Given: Two concentric circles C₁ and C₂ with a common center O. A chord AB of the larger circle C₁ touches the smaller circle C₂ at a point P.

To Prove: The chord AB is bisected at the point of contact P, i.e., AP = BP.

{{VISUAL: diagram: Two concentric circles with center O. A horizontal chord AB of the larger circle is drawn such that it is tangent to the smaller circle at point P. A radius OP is drawn from O to P, which is perpendicular to AB.}}

Solution:

1. Identify the relationship: The line segment AB is a tangent to the smaller circle C₂ at the point P. The line segment OP is the radius of the smaller circle C₂.

2. Apply Theorem 10.1: We know that the radius is perpendicular to the tangent at the point of contact.

   OP ⊥ AB
   

3. Consider the larger circle: Now, let's focus on the larger circle C₁. Here, AB is a chord and OP is a perpendicular line drawn from the center O to this chord.

4. Recall chord properties: A theorem from earlier studies states that the perpendicular drawn from the center of a circle to a chord bisects the chord.

5. Conclusion: Since OP ⊥ AB, OP must bisect the chord AB. This directly implies that the point P divides the chord AB into two equal halves.

   AP = BP
   

Final Answer: Hence, it is proved that the chord of the larger circle is bisected at the point of contact with the smaller circle.

Example 3: Angle Relationship in Tangents (Hard)

Given: Two tangents TP and TQ are drawn to a circle with center O from an external point T. P and Q are the points of contact.

To Prove: ∠PTQ = 2 ∠OPQ

{{VISUAL: diagram: A circle with center O. An external point T. Two tangents TP and TQ are drawn from T touching the circle at P and Q. Lines OP, OQ, and OT are drawn. The angle ∠PTQ is labeled θ.}}

Solution:

1. Let's denote ∠PTQ by the variable θ.

   ∠PTQ = θ
   

2. By Theorem 10.2, the lengths of tangents from an external point are equal. Thus, TP = TQ. This makes ΔTPQ an isosceles triangle.

3. In an isosceles triangle, angles opposite to equal sides are equal. Therefore, ∠TPQ = ∠TQP. The sum of angles in a triangle is 180°.

   ∠TPQ = ∠TQP = (180° - θ) / 2 = 90° - θ/2
   

4. By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. So, OP ⊥ TP, which means ∠OPT = 90°.

5. The angle ∠OPT is made up of two smaller angles, ∠OPQ and ∠TPQ. So, ∠OPT = ∠OPQ + ∠TPQ. We can write ∠OPQ = ∠OPT - ∠TPQ.

6. Substitute the values we know into this equation.

   ∠OPQ = 90° - (90° - θ/2)
   

7. Simplify the expression.

   ∠OPQ = 90° - 90° + θ/2
   

   ∠OPQ = θ/2
   

8. Since we defined θ = ∠PTQ, we can substitute it back.

   ∠OPQ = (∠PTQ) / 2
   

9. Rearranging this gives the desired result.

   ∠PTQ = 2 ∠OPQ
   

Final Answer: Hence, it is proved that ∠PTQ = 2 ∠OPQ.

Example 4: Finding the Tangent Length (Tricky)

Given: PQ is a chord of length 8 cm of a circle with a radius of 5 cm. The tangents at P and Q intersect at an external point T.

To Find: The length of the tangent TP.

{{VISUAL: diagram: A circle with center O. A vertical chord PQ is drawn. Tangents from P and Q meet at an external point T. The line OT intersects the chord PQ at R. Lengths are labeled: OP=5, PR=4. OT is the angle bisector of ∠PTQ.}}

Solution:

1. Construction: Join OT. Let OT intersect the chord PQ at point R.

2. Symmetry Properties: In ΔTPQ, TP = TQ (tangents from an external point). So, ΔTPQ is an isosceles triangle. The line TO is the angle bisector of ∠PTQ and is also the perpendicular bisector of the chord PQ.

3. Chord Bisection: Therefore, OT ⊥ PQ and R is the midpoint of PQ.

   PR = RQ = PQ / 2 = 8 / 2 = 4 cm
   

4. Find OR: Now consider the right-angled triangle ΔOPR. The hypotenuse OP is the radius (5 cm) and PR is 4 cm. We can find OR using Pythagoras theorem.

   OR² + PR² = OP²
   

   OR² + 4² = 5²
   

   OR² + 16 = 25
   

   OR² = 9  =>  OR = 3 cm
   

5. Using Similar Triangles: Consider the right-angled triangles ΔTRP and ΔPRO.

   • ∠TRP = ∠PRO = 90° (Proved in Step 2, but ∠PRO is a straight line with ∠TRP). Let's rethink this. OT ⊥ PQ, so ∠PRT = 90°. ∠PRO = 90°. So this is wrong. Let's re-read the NCERT logic.
   • Ah, the NCERT logic is: In ΔOPT, ∠OPT = 90°. So ∠TPR + ∠RPO = 90°. In ΔTRP, ∠TPR + ∠PTR = 90°. Comparing these gives ∠RPO = ∠PTR.

6. Establishing Similarity: Now, in ΔTRP and ΔPRO:

   • ∠TRP = ∠PRO = 90° (Wait, this is wrong in the diagram. ∠TRP is 90°, ∠ORP is not necessarily 90°). Let's follow the NCERT's logic exactly. ∠TRP = 90° and ∠ORP = 90°. Both are correct. Let's check ΔPRO again. Yes, ∠ORP = 90° is what we found OT ⊥ PQ. My mistake. So ∠TRP should be ∠PRT which is 90°.
   • Ah, the NCERT says right triangle TRP is similar to the right triangle PRO. Let's verify the angles.
   • In ΔTRP: Angles are ∠PTR, ∠TPR, ∠PRT (90°).
   • In ΔPRO: Angles are ∠POR, ∠OPR (same as ∠RPO), ∠ORP (90°).
   • We know ∠PTR = ∠OPR (from the logic in step 5).
   • Since both have a 90° angle and one other angle is equal, the third angles must also be equal. So, ∠TPR = ∠POR.
   • Therefore, by AA similarity, ΔTRP ~ ΔPRO.

7. Using Ratios of Similar Triangles: The ratio of corresponding sides must be equal.

   TP / PO = RP / RO = TR / PR
   

8. Let's use the first two parts of the ratio to find TP.

   TP / PO = RP / RO
   

   TP / 5 = 4 / 3
   

9. Solve for TP.

   TP = (4 × 5) / 3 = 20/3
   

Final Answer: The length of TP is 20/3 cm.

Tips & Tricks

Use these properties to solve problems faster, especially in multiple-choice questions.

Common Mistakes

Be careful to avoid these common errors when solving problems related to tangents.

Brain-Teaser Questions

Test your understanding with these challenging problems.

1. A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC, CD, and DA at points P, Q, R, and S respectively. If AB = 6 cm, BC = 7 cm, and CD = 4 cm, find the length of AD.

   💡 Answer: We use the property that tangents from an external point are equal. AP = AS (from point A) BP = BQ (from point B) CR = CQ (from point C) DR = DS (from point D) Adding them all: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC 6 + 4 = AD + 7 10 = AD + 7 => AD = 3 cm.

2. Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the larger circle which touches the smaller circle.

   💡 Answer: This is the reverse of Example 2. Let the chord of the larger circle be AB, touching the smaller circle at P. O is the center. OA = 13 cm (radius of big circle), OP = 5 cm (radius of small circle). OP ⊥ AB. In right ΔOPA, AP² + OP² = OA². AP² + 5² = 13² AP² + 25 = 169 AP² = 144 => AP = 12 cm. Since the perpendicular from the center bisects the chord, AB = 2 × AP = 2 × 12 = 24 cm.

3. From an external point P, two tangents PA and PB are drawn to a circle with center O. If the angle ∠APB = 120°, prove that OP = 2 AP.

   💡 Answer: Join OP, OA. OP bisects ∠APB, so ∠APO = 120° / 2 = 60°. In right-angled ΔOAP (right angle at A), we can use trigonometry. cos(∠APO) = Adjacent / Hypotenuse cos(60°) = AP / OP We know cos(60°) = 1/2. 1/2 = AP / OP Cross-multiplying gives OP = 2 AP.

Mini Cheatsheet

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Summary & Quick Revision

Page 5 of 5: Summary & Quick Revision

Welcome to the final revision page for our chapter on Circles! We've journeyed through the elegant properties of tangents and their relationship with the circle's radius. Now, it's time to consolidate that knowledge, ensuring these concepts are locked in for your exams and for your future mathematical explorations. This summary is designed to be your quick-reference guide, highlighting the most critical theorems and problem-solving techniques. Let's sharpen our understanding and master the geometry of circles together.

Concept Introduction: Circles in the Real World

Have you ever looked at the belt system in a car engine or a complex pulley system on a crane? The flat, straight parts of the belt that run between the circular pulleys are perfect real-world examples of tangents. The belt touches each pulley at exactly one point before moving on to the next. The designers of these systems rely on the geometric principles we've learned. They need to calculate the exact length of the belt, which involves knowing the radius of the pulleys and the lengths of the tangent sections. This ensures the belt has the correct tension to run the engine's components smoothly. This is a beautiful illustration of how abstract geometric rules govern the mechanics of the world around us.

{{FORMULA: expr=PA = PB | symbols=P:External Point, A:Point of Contact 1, B:Point of Contact 2}}

Key Definitions & Theorems

This chapter revolves around a few core ideas. Let's formalize them in a table.

Derivation: Proof of Theorem 10.2

Let's walk through the logical proof for one of the most powerful theorems in this chapter: The lengths of tangents drawn from an external point to a circle are equal.

Statement: Given a circle with center O and an external point P. Two tangents, PA and PB, are drawn from P to the circle, touching the circle at points A and B respectively. We need to prove that PA = PB.

{{VISUAL: diagram: A circle with center O. An external point P. Two tangents PA and PB are drawn from P to the circle at points of contact A and B. Lines OA, OB, and OP are drawn as dotted lines, forming two right-angled triangles ΔOAP and ΔOBP.}}

Construction: Join the center O to the external point P, and also join O to the points of contact A and B. This gives us two triangles: ΔOAP and ΔOBP.

Proof:

1. Identify the known properties of the two triangles, ΔOAP and ΔOBP. We know that the radius is perpendicular to the tangent at the point of contact (Theorem 10.1).

   ∠OAP = 90°
   ∠OBP = 90°
   

2. Observe the sides of the two triangles. OA and OB are radii of the same circle, so they must be equal.

   OA = OB (Radii of the same circle)
   

3. The side OP is common to both triangles.

   OP = OP (Common side)
   

4. We now have two right-angled triangles where the hypotenuse (OP) and one side (OA and OB) are equal. We can apply the Right angle-Hypotenuse-Side (RHS) congruence criterion.

   ΔOAP ≅ ΔOBP (By RHS congruence)
   

5. Since the triangles are congruent, their corresponding parts must be equal (CPCTC - Corresponding Parts of Congruent Triangles are Congruent).

   PA = PB
   

Thus, we have successfully proven that the lengths of the two tangents from an external point to a circle are equal. This property is fundamental to solving a majority of problems in this chapter.

Solved Examples

Let's apply these theorems to solve some problems, starting from easy and moving to more complex ones.

Example 1: Finding a Tangent's Length (Easy)

Given: A point P is 13 cm from the center O of a circle. The radius of the circle is 5 cm. A tangent is drawn from P to the circle at point T.

To Find: The length of the tangent PT.

Solution:

1. According to Theorem 10.1, the radius OT is perpendicular to the tangent PT at the point of contact T. Therefore, ΔOTP is a right-angled triangle with the right angle at T.

   ∠OTP = 90°
   

2. The hypotenuse of this triangle is the line joining the center to the external point, which is OP. We can apply the Pythagoras theorem: OT² + PT² = OP².

   5² + PT² = 13²
   

3. Now, we solve for PT.

   25 + PT² = 169
   

4. Isolate PT² and find the square root.

   PT² = 169 - 25 = 144
   

   PT = √144 = 12 cm
   

Final Answer: The length of the tangent PT is 12 cm.

Example 2: Quadrilateral Circumscribing a Circle (Medium)

Given: A quadrilateral ABCD is drawn to circumscribe a circle.

To Find: Prove that AB + CD = AD + BC.

{{VISUAL: diagram: A circle is inside a quadrilateral ABCD. The circle touches the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.}}

Solution:

1. Let the points where the circle touches the sides AB, BC, CD, and DA be P, Q, R, and S, respectively.

2. We use Theorem 10.2, which states that tangents from an external point to a circle are equal in length. Let's apply this for each vertex (external point) of the quadrilateral.

   • From point A: AP = AS (Equation 1)
   • From point B: BP = BQ (Equation 2)
   • From point C: CR = CQ (Equation 3)
   • From point D: DR = DS (Equation 4)

3. Now, we add all four equations together.

   (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
   

4. By observing the figure, we can group the segments to form the sides of the quadrilateral. AP + BP = AB, CR + DR = CD, AS + DS = AD, and BQ + CQ = BC.

   AB + CD = AD + BC
   

Final Answer: Hence, it is proved that for a quadrilateral circumscribing a circle, the sum of opposite sides is equal.

Example 3: Concentric Circles Problem (Hard)

Given: Two concentric circles have radii 5 cm and 3 cm. A chord of the larger circle touches the smaller circle.

To Find: The length of this chord.

Solution:

1. Let the two concentric circles have a common center O. Let the larger circle be C1 and the smaller circle be C2. Let AB be the chord of C1 that touches C2 at point P.

   {{VISUAL: diagram: Two concentric circles with center O. A chord AB of the larger circle is shown. This chord is tangent to the smaller circle at a point P. Radii OA (to the larger circle) and OP (to the smaller circle) are drawn, forming a right-angled triangle OPA.}}

2. Since AB is tangent to the smaller circle C2 at point P, the radius OP of C2 is perpendicular to the chord AB (Theorem 10.1).

   OP ⊥ AB
   

3. This means that ΔOPA is a right-angled triangle, with the right angle at P. We are given the lengths of the hypotenuse OA (radius of C1) and the side OP (radius of C2).

   • OA = 5 cm
   • OP = 3 cm

4. Apply the Pythagoras theorem to find the length of AP.

   OP² + AP² = OA²
   

   3² + AP² = 5²
   

   9 + AP² = 25
   

   AP² = 16  =>  AP = 4 cm
   

5. A perpendicular from the center of a circle to a chord bisects the chord. Since OP ⊥ AB, P is the midpoint of AB.

   AB = 2 × AP
   

   AB = 2 × 4 = 8 cm
   

Final Answer: The length of the chord is 8 cm.

{{KEY: type=concept | title=Combining Theorems | text=Notice how the hardest problems often require you to use both major theorems together. The radius is perpendicular to the tangent (Theorem 10.1), which often creates a right-angled triangle, allowing you to use Pythagoras Theorem. This is often combined with the equality of tangents from an external point (Theorem 10.2).}}

Example 4: Angles between Tangents (Tricky)

Given: Two tangents TP and TQ are drawn to a circle with center O from an external point T.

To Find: Prove that ∠PTQ = 2∠OPQ.

Solution:

1. Let the angle between the two tangents be ∠PTQ. We know from Theorem 10.2 that TP = TQ. Therefore, ΔTPQ is an isosceles triangle.

   ∠TQP = ∠TPQ
   

2. The sum of angles in ΔTPQ is 180°. So, ∠PTQ + ∠TQP + ∠TPQ = 180°. Substituting ∠TQP with ∠TPQ, we get:

   ∠PTQ + 2∠TPQ = 180°
   

   2∠TPQ = 180° - ∠PTQ
   

   ∠TPQ = ½ (180° - ∠PTQ) = 90° - ½∠PTQ   (Equation 1)
   

3. From Theorem 10.1, the radius OP is perpendicular to the tangent TP.

   ∠OPT = 90°
   

4. From the figure, we can see that ∠OPT = ∠OPQ + ∠TPQ.

   90° = ∠OPQ + ∠TPQ
   

   ∠TPQ = 90° - ∠OPQ   (Equation 2)
   

5. Now, we equate the expressions for ∠TPQ from Equation 1 and Equation 2.

   90° - ½∠PTQ = 90° - ∠OPQ
   

6. Simplifying the equation by cancelling 90° from both sides and multiplying by -1.

   ½∠PTQ = ∠OPQ
   

   ∠PTQ = 2∠OPQ
   

Final Answer: Hence, it is proved that the angle between the two tangents is twice the angle between the chord connecting the points of contact and the radius to one of the points of contact.

Tips & Tricks

Common Mistakes

Avoiding these common pitfalls will significantly improve your accuracy.

Brain-Teaser Questions

1. Supplementary Angles at the Center: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (This is Q13 from your NCERT book!)

   💡 Answer: Let the quadrilateral be ABCD circumscribing a circle with center O, touching the sides at P, Q, R, S. Join OA, OB, OC, OD. In ΔOAP and ΔOAS, they are congruent by SSS/RHS. So ∠AOP = ∠AOS. Similarly, ∠BOP = ∠BOQ, etc. Let these half angles be a, a, b, b, c, c, d, d. The sum of all angles around O is 360°, so 2a + 2b + 2c + 2d = 360°, which means a+b+c+d = 180°. The angle subtended by side AB is ∠AOB = a+b and by side CD is ∠COD = c+d. Their sum is (a+b) + (c+d) = 180°. Similarly, ∠BOC + ∠DOA = 180°.

2. Incircle of a Right Triangle: A right-angled triangle ABC with sides 6 cm, 8 cm, and 10 cm circumscribes a circle. What is the radius of the circle?

   💡 Answer: Let the sides be a=8, b=6, c=10. The radius r of an incircle of a right-angled triangle can be found using the formula: r = (a + b - c) / 2. Here, r = (8 + 6 - 10) / 2 = 4 / 2 = 2 cm. Alternatively, you can prove this by equating the area of the triangle (½ × base × height) to the sum of the areas of the three small triangles formed by joining vertices to the incenter (½ × r × (a+b+c)).

3. Two Touching Circles: Two circles with radii 5 cm and 3 cm touch each other externally. What is the length of their direct common tangent?

   💡 Answer: Let the centers be O1 and O2, and radii be r1=5 and r2=3. Let the direct common tangent be AB. The distance between centers O1O2 = r1 + r2 = 8 cm. If you draw a line from O2 parallel to AB, meeting the radius O1A at a point C, you form a right-angled triangle O1CO2. Here, O1O2 is the hypotenuse (8 cm), and O1C = r1 - r2 = 5 - 3 = 2 cm. The length of the tangent AB is equal to CO2. Using Pythagoras: CO2² = O1O2² - O1C² = 8² - 2² = 64 - 4 = 60. The length of the tangent is √60 = 2√15 cm.

Mini Cheatsheet

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